millennium force: cedar point roller coaster simulation...ef151q@ef.utk.edu ef 151 fall, 2015...

EF 151 Fall, 2015 Lecture 3-3 1

A 2800 lb sports car accelerates on level ground from rest to 60 mph in

7.4 seconds. What is the maximum power output required (in hp)?

Ignore friction and air resistance. 1 hp = 550 ft-lb/sec

Review Problem – Power and Acceleration

Today’s Topics:

� Review: Power� Roller Coasters� Springs� Elastic Potential Energy

ef151q@ef.utk.eduEF 151 Fall, 2015 Lecture 3-3 2

Millennium Force: Cedar Point

What is theoretical speed at the bottom of the 310 ft drop (assuming no initial velocity at top of hill)?A. 90 mphB. 92 mphC. 96 mph

What is the theoretical speed at the bottom of the hill if the speed at the top is 20 mph?A. 96 mphB. 98 mphC. 116 mph

Is the theoretical speed at the bottom of the hill a function of how steep the hill is?

Is the theoretical speed at the bottom of the hill a function of the weight of the coaster?

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EF 151 Fall, 2015 Lecture 3-3 3

Millennium Force: Cedar Point

The actual speed is reported as 92 mph. What is the efficiency of the Millennium Force (assuming a speed at the top of 20mph)?

ef151q@ef.utk.eduEF 151 Fall, 2015 Lecture 3-3 4

Roller Coaster Simulation

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EF 151 Fall, 2015 Lecture 3-3 5

Linear Springs

� Force is proportional to deflection of spring from “_____________”

� F is force in spring

� x is the deflection of the spring from “____________”

� Spring constant, k

� units of ______ per unit

_______________

� N/m, lb/ft

kxF =

F

x

k

ef151q@ef.utk.eduEF 151 Fall, 2015 Lecture 3-3 6

Elastic Potential Energy

� Within a certain range of motion, linear springs are essentially _______________

� work put into the spring is available as energy out of the spring at a later time

� Only _________ and _________ position are necessary

� Uelas = ___________

k

fF

Δx0

Work = Area under F-Δx curve

W = ½ FΔx

W = ½ (kΔx)Δx

W = ½ k(Δx)2

Units:ef151q@ef.utk.edu

EF 151 Fall, 2015 Lecture 3-3 7

Conservation of Mechanical Energy

K0 + Ugrav0 + Uelas0 + Win = Kf + Ugravf + Uelasf + Eloss

• 0 – initial state

• f – final state

• h measured from ____________

• x measured from the ____________ position

• Win – positive value representing all work put into the system

• Eloss – positive value representing all energy “lost”

ef151q@ef.utk.eduEF 151 Fall, 2015 Lecture 3-3 8

Object Launched by a Spring-Gun

Given:

• 1/8 lb ball is shot straight upward

• k = 40 lb/ft

• Undeformed length of spring is 1.2 ft

Required:

• Force required to deform the spring

• Height above ground that ball goes

• Speed of the ball when it leaves the spring

1.2

ft

0.7

ft

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EF 151 Fall, 2015 Lecture 3-3ef151q@ef.utk.edu

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EF 151 Fall, 2015 Lecture 3-3ef151q@ef.utk.edu

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